Problem Sheet 1

Section B.1 Problem Sheet 1

Due 4 October at 3pm, either in class or in the pigeon hole in 4W.

Exercise B.1. Quantifiers and limits.

Let \(\seq an\) be a sequence of real numbers and let \(L \in \R\text{.}\)

(a)

Write the precise definition of “\(a_n \to L\) as \(n\to \infty\)”. Spell out any quantifiers \(\forall,\exists\) as words.

Solution.

For any \(\varepsilon \gt 0\text{,}\) there exists \(N \in \N\) such that, for all \(n \ge N\text{,}\) \(\abs{a_n - L} \lt \varepsilon\text{.}\)

(b)

Which of the following is the logical negation of the statement “\(a_n \to L\) as \(n \to \infty\)”?

For all \(\varepsilon \gt 0\text{,}\) and for all \(N \in \N\text{,}\) there exists \(n \ge N\) such that \(\abs{a_n-L} \ge \varepsilon\text{.}\)
There exists \(\varepsilon \gt 0\) such that, for all \(N \in \N\text{,}\) there exists \(n \ge N\) such that \(\abs{a_n-L} \ge \varepsilon\text{.}\)
There exists \(\varepsilon \gt 0\) and \(N \in \N\) such that, for all \(n \ge N\text{,}\) \(\abs{a_n-L} \ge \varepsilon\text{.}\)
For all \(\varepsilon \gt 0\text{,}\) there exists \(N \in \N\) such that, for all \(n \ge N\text{,}\) \(\abs{a_n-L} \ge \varepsilon\text{.}\)

Solution.

The logical negation of “\(a_n \to L\) as \(n \to \infty\)” is (ii).

Comment.

Being able to quickly and accurately negate statements involving quantifiers is a basic skill in Analysis. In principle there are rules that one can memorise, but in more complicated situations it is also important to have internalised why these rules make sense. If you had any difficulty with this question whatsoever, I encourage you to look through any relevant notes from Analysis 1 and practice negating other statements like “the function \(f \maps (a,b) \to \R\) is continuous at the point \(x_0 \in (a,b)\)”. I am also more than happy to talk about this in office hours.

(c)

Let \(\seq an\) and \(\seq bn\) be sequences of real numbers with \(\lim_{n \to \infty} a_n = 0\) and \(\lim_{n \to \infty} b_n = 1\text{.}\) The conclusion of the argument below is clearly incorrect, but where does the reasoning go wrong? How could the argument have been written to make such a mistake less likely? Note that I am not asking you to give a correct proof of a related true statement.

For any \(\varepsilon_1 \gt 0\text{,}\) we can find \(N_1 \in \N\) such that \(\abs{a_n} \lt \varepsilon_1\) for all \(n \ge N_1\text{.}\) Similarly, for any \(\varepsilon_2 \gt 0\) we can find \(N_2 \in \N\) such that \(\abs{b_n-1} \le \varepsilon_2\) for \(n \ge N_2\text{.}\) Let \(N=\max\{N_1,N_2\}\text{.}\) Then for \(n \ge N\text{,}\)

\begin{align*} \abs{a_n+b_n} \amp = \abs{a_n+(b_n-1)+1}\\ \amp \le \abs{a_n}+\abs{b_n-1}+1\\ \amp \lt \varepsilon_1 + \varepsilon_2 + 1\\ \amp = \varepsilon\text{.} \end{align*}

Therefore \(\lim_{n\to \infty}(a_n + b_n)=0\text{.}\)

Solution.

See Section G.1.

Comment.

This question asks you to explain what was wrong with this incorrect argument (of a false statement). It did not ask you to give a correct argument (of a related true statement). Explaining why an argument doesn’t work can often be quite tricky, and is a useful mathematical skill to develop.

Exercise B.2. Suprema and limits.

Let \(A \subset \R\) be a nonempty set.

(a)

Write down the definition of \(\sup A\) in complete detail.

Hint.

If you have any doubts, look it up in your old notes!

Solution.

We call \(s \in \R \cup \{\infty\}\) an upper bound for \(A\) if \(a \le s\) for all \(a \in A\text{.}\) The supremum \(\sup A \in \R \cup \{\infty\}\) is the least upper bound for \(A\) in the sense that

it is an upper bound for \(A\text{;}\) and
if \(s\) is an upper bound for \(A\text{,}\) then \(s \ge \sup A\text{.}\)

Comment 1.

Occasionally I will see the more compact definition

\begin{equation*} \sup A = \min\set{x \in \R}{x \ge a \text{ for all } a \in A}\text{.} \end{equation*}

This is of course equivalent to the definition given in these solutions, provided you have a clear understanding of what is meant by \(\min\). For instance, in order for this definition to work we need to agree that the minimum of the empty set is \(+\infty\text{.}\) If you find working with suprema confusing, I would encourage to use the longer definition in these solutions.

Comment 2.

The second statement in the official solution is equivalent to

for all \(\varepsilon \gt 0\) there exists \(a \in A\) such that \(a \gt \sup A - \varepsilon\text{,}\)

and so we could have written that instead.

(b)

Suppose that \(\sup A \lt \infty\text{.}\) Using the above definition, show that there exists a sequence \(\seq an\) of points in \(A\) such that \(a_n \to \sup A\) as \(n \to \infty\text{.}\)

Hint.

Let \(n \in \N\text{.}\) Is \(\sup A - 1/n\) an upper bound for \(A\text{?}\)

Solution.

For each \(n \in \N\text{,}\) we know from the above definition that \(\sup A - 1/n\) is not an upper bound for \(A\text{.}\) Thus there must exist \(a_n \in A\) with \(\sup A - 1/n \lt a_n \le \sup A\text{.}\) In particular, \(\abs{a_n-\sup A} \le 1/n \to 0\) as \(n \to \infty\text{,}\) which implies \(a_n \to \sup A\text{.}\)

Comment 1.

Note that there is no reason whatsoever for \(\sup A - 1/n\) to be an element of \(A\text{.}\) For instance, if \(\sup A = \{0\}\) is a set with a single element, then this is never the case.

Comment 2.

There is no guarantee whatsoever that a given infinite set will have a minimum (or a maximum). For instance,

\begin{equation*} \min\set{x \in \R}{x \gt 0} \end{equation*}

does not exist. But

\begin{equation*} \inf\set{x \in \R}{x \gt 0} = 0 \end{equation*}

does exist. This is an important distinction which will come up many times in the unit. If it is at all mysterious to you, please take a look at your old Analysis 1 notes and/or come talk to me in office hours.

(c)

Give counterexamples to show that each of the arguments below is incorrect.

Suppose that every \(a \in A\) satisfies \(a \lt \infty\text{.}\) Then \(\sup A \lt \infty\text{.}\)
Suppose that every \(a \in A\) satisfies \(a \lt 1\text{.}\) Then \(\sup A \lt 1\text{.}\)

Hint.

Both arguments are valid when \(A\) is a finite set, in which case \(\sup\) can be replaced by \(\max\text{.}\) To find counterexamples, you will therefore need to think about infinite sets \(A\text{.}\)

Solution.

For the first argument, the hypothesis is always true since all real numbers \(x \in \R\) satisfy \(x \lt \infty\text{.}\) But clearly the conclusion is not always true. For instance, we could take \(A = \N\text{.}\)

For the second argument, a counterexample is the open interval \(A = (0,1)\text{.}\) Then \(a \lt 1\) for all \(a \in A\text{,}\) but \(\sup A = 1\) is not \(\lt 1\text{.}\)

Comment 1.

Note that the previous part of this question is not and “if and only if”. Let \(\seq an\) be a sequence of points in \(A\) which converges to some real number \(L\text{.}\) It is a good Analysis 1 exercise to show that this implies the inequality \(\sup A \ge L\text{.}\) On the other hand it is easy to cook up examples where this inequality is strict, i.e. where \(\sup A \gt L\text{.}\)

Comment 2.

Many students, instead of giving set \(A \subset \R\text{,}\) gave a sequence \(\seq an\) of real numbers. I assume that these students meant to take the set \(A = \{a_n : n \in \N\}\) of all points in this sequence, but in this unit this is something you have to spell out. Also see Section G.3.

Exercise B.3. Inequalities and limits.

Let \(\seq an\text{,}\) \(\seq bn\) be convergent sequences of real numbers, with \(a_n \to a\) and \(b_n \to b\) as \(n \to \infty\text{.}\)

(a)

Directly using the definition of the limit of a sequence, show that if \(a \lt b\text{,}\) then there exists \(N \in \N\) such that \(a_n \lt b_n\) for all \(n \ge N\text{.}\)

Hint.

Try sketching the sequences and their limits on a number line to get inspiration.

Solution 1.

Let \(\varepsilon = (b-a)/2\text{.}\) Then using the definition of convergence there exists \(N \in \N\) such that, for all \(n \ge N\text{,}\) \(\abs{a_n-a} \lt \varepsilon\) and \(\abs{b_n - b} \lt \varepsilon\text{.}\) Thus, for \(n \ge N\text{,}\)

\begin{align*} b_n - a_n \amp = (b_n-b) + (b-a) + (a-a_n)\\ \amp \gt -\varepsilon + 2\varepsilon - \varepsilon\\ \amp = 0 \end{align*}

as desired.

Solution 2. Motivation for the solution

Here is a more rambling explanation of one way to come up with the official solution. Fix some \(\varepsilon \gt 0\) to be determined. Using the definition of convergence twice, we can find \(N \in \N\) such that, for all \(n \ge N\text{,}\) \(\abs{a_n-a} \lt \varepsilon\) and \(\abs{b_n-b} \lt \varepsilon\text{.}\) It is useful to draw a picture of what these two inequalities look like on a number line. In principle we could have used a different \(\varepsilon\) for each sequence, but it turns out that using the same one is good enough.

Suppose that \(n \ge N\text{.}\) What do these bounds on \(\abs{a_n-a}\) and \(\abs{b_n-b}\) tell us? We want to show that \(a_n \lt b_n\text{,}\) and so it would be nice to have an upper bound for \(a_n\) and a lower bound for \(b_n\text{.}\) The inequalities that we already have are

Alternatively, we could write \(a_n = a + (a_n - a)\) leading to the estimate \(a_n \le a + \abs{a_n-a} \lt a + \varepsilon\text{,}\) and something similar for \(b_n\text{.}\)

\begin{equation*} -\varepsilon \lt a_n - a \lt \varepsilon \quad \anda \quad -\varepsilon \lt b_n - b \lt \varepsilon\text{,} \end{equation*}

which after some manipulations imply

\begin{equation*} a_n \lt a + \varepsilon \quad \anda \quad b_n \gt b - \varepsilon\text{.} \end{equation*}

These bounds point in the correct direction, but do they imply that \(a_n \lt b_n\text{?}\) They would if our upper bound for \(a_n\) was \(\le\) our lower bound for \(b_n\text{,}\) i.e. if

\begin{equation*} a + \varepsilon \le b - \varepsilon. \end{equation*}

Rearranging this inequality, we see that things will work out provided we pick

\begin{equation*} \varepsilon \le \frac{b-a}2, \end{equation*}

which we’re allowed to do since \((b-a)/2 \gt 0\text{.}\)

Having worked through all of this on scratch paper, we are now ready to pull out a fresh sheet of paper and write up the argument, choosing \(\varepsilon \gt 0\) at the very start. Any positive \(\varepsilon \le (b-a)/2\) will do, but for simplicity we pick \(\varepsilon = (b-a)/2\text{.}\) As often happens, this clean argument is much shorter! Alternatively, if we are very careful, we could write up a version of the argument which leaves \(\varepsilon\) to be determined at the start and then chooses a value of \(\varepsilon\) at the end, without falling into the trap discussed in Section G.1.

(b)

Find an example where \(a \le b\text{,}\) but \(a_n \gt b_n\) for all \(n \in \N\text{.}\)

Solution.

We could take, for instance, \(a_n = 1/n\) and \(b_n = 0\) for all \(n \in \N\text{.}\) Then clearly \(a = b = 0\text{,}\) and so \(a \le b\text{,}\) but \(a_n \gt b_n \) for all \(n \in \N\text{.}\)

(c)

Directly using the definition of the limit of a sequence, show that if \(a_n \le b_n\) for all \(n \in \N\text{,}\) then \(a \le b\text{.}\)

Hint.

Try proving the contrapositive.

Solution.

We prove the contrapositive. Suppose that \(a \gt b\text{.}\) Then by Part a (with the roles of \(a\) and \(b\) reversed), there exists \(N \in \N\) such that \(a_n \gt b_n\) for all \(n \ge N\text{.}\) In particular, we have \(a_N \gt b_N\text{.}\)

(d)

Find an example where \(a_n \lt b_n\) for all \(n \in \N\) but \(a \ge b\text{.}\)

Solution.

We can take \(a_n = 0\) and \(b_n = 1/n\text{.}\) Clearly \(a_n \lt b_n\) for all \(n \in \N\text{,}\) but \(a = b = 0\text{.}\)

While Chapter 1 begins with a review of Analysis 2A, my experience has been that many students would also benefit from a review of Analysis 1. So, if you are feeling at all rusty with Analysis 1 techniques, I encourage you in the strongest possible terms to attempt some of the questions on this problem sheet and turn them in so that you can get feedback.

Please feel free to email me, or drop by office hours, which in Week 1 are Wednesday 10:15–11:05 in 4W 1.12, with any questions about these problems whatsoever. I am also more than happy to meet one-on-one or in a small group.

In particular, if these problems are very difficult for you, please get in touch with me over email or in office hours as soon as you can. It is possible that you will need to do a more thorough revision of Analysis 1 and/or that this unit may not be a great fit for you.

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